Unit Systems and Conversions – Complete Guide
1. Classification of Unit Systems (Exercise 1.1.01)
Necessity of Units
All physical quantities are to be measured in terms of standard quantities.
Unit Definition
A unit is defined as a standard or fixed quantity of one kind used to measure other quantities of the same kind.
Classification of Units
Units are classified into two main categories:
- Fundamental units: Units of basic quantities of length, mass and time.
- Derived units: Units which are derived from basic units and bear a constant relationship with the fundamental units (e.g., area, volume, pressure, force).
Systems of Units
| System | Description | Length | Mass | Time |
|---|---|---|---|---|
| F.P.S | British system | Foot (ft) | Pound (lb) | Second (s) |
| C.G.S | Metric system | Centimeter (cm) | Gram (g) | Second (s) |
| M.K.S | Metric system | Metre (m) | Kilogram (kg) | Second (s) |
| S.I. | International System | Metre (m) | Kilogram (kg) | Second (s) |
Example Problem: Length Calculation
What is the length of copper wire in the roll, if the roll of copper wire weighs 8kg, the diameter of wire is 0.9cm and the density is 8.9 gm/cm³?
Solution:
Given:
- Mass of copper wire = 8 kg = 8000 grams
- Diameter of copper wire = 0.9 cm
- Density of copper wire = 8.9 gm/cm³
Step 1: Calculate cross-sectional area
Area = πd²/4 = π(0.9)²/4 = 0.636 cm²
Step 2: Calculate volume
Volume = Mass/Density = 8000g / 8.9 g/cm³ = 898.88 cm³
Step 3: Calculate length
Length = Volume/Area = 898.88 cm³ / 0.636 cm² = 1413.33 cm
Final Answer: Length of copper wire = 1413 cm
2. Fundamental and Derived Units (Exercise 1.1.02)
Fundamental Units Comparison
| S.No. | Basic quantity | British units (F.P.S) | Metric units (C.G.S) | Metric units (M.K.S) | International units (S.I.) |
|---|---|---|---|---|---|
| 1 | Length | Foot (ft) | Centimetre (cm) | Metre (m) | Metre (m) |
| 2 | Mass | Pound (lb) | Gram (g) | Kilogram (kg) | Kilogram (kg) |
| 3 | Time | Second (s) | Second (s) | Second (s) | Second (s) |
| 4 | Current | Ampere (A) | Ampere (A) | Ampere (A) | Ampere (A) |
| 5 | Temperature | Fahrenheit (°F) | Centigrade (°C) | Centigrade (°C) | Kelvin (K) |
| 6 | Light intensity | Candela (Cd) | Candela (Cd) | Candela (Cd) | Candela (Cd) |
Derived Units Comparison
| S.No. | Physical quantity | British units (FPS) | Metric units (CGS) | Metric units (MKS) | International units (SI) |
|---|---|---|---|---|---|
| 1 | Area | Square foot (ft²) | Square centimetre (cm²) | Square metre (m²) | Square metre (m²) |
| 2 | Volume | Cubic foot (ft³) | Cubic centimetre (cm³) | Cubic metre (m³) | Cubic metre (m³) |
| 3 | Density | Pound per cubic foot (lb/ft³) | Gram per cubic centimetre (g/cm³) | Kilogram per cubic metre (kg/m³) | Kilogram per cubic metre (kg/m³) |
| 4 | Speed | Foot per second (ft/s) | Centimetre per second (cm/s) | Metre per second (m/s) | Metre per second (m/s) |
| 13 | Pressure, Stress | Pound per square inch (lb/in²) | Gram per square centimetre (g/cm²) | Kilogram per square metre (kg/m²) | Newton per square metre (N/m²) |
| 22 | Moment of force | Pounds foot (lbs.ft) | Gram centimetre (g.cm) | Kilogram metre (kg.m) | Newton metre (Nm) |
Note: The complete table includes 36 physical quantities with their units in all four systems. The above shows a sample of the most commonly used units.
3. Measurement Units and Conversion (Exercise 1.1.03)
Decimal Multiples and Submultiples
| Decimal power | Value | Prefix | Symbol |
|---|---|---|---|
| 1012 | 1,000,000,000,000 | tera | T |
| 109 | 1,000,000,000 | giga | G |
| 106 | 1,000,000 | mega | M |
| 103 | 1,000 | kilo | k |
| 10-3 | 0.001 | milli | m |
| 10-6 | 0.000001 | micro | μ |
Units and Abbreviations
| Quantity | Units | Abbreviation |
|---|---|---|
| Length | millimetre, metre, kilometre | mm, m, km |
| Mass | kilogram, gram | kg, g |
| Time | seconds, minutes, hours | s, min, h |
| Speed | metre per second, kilometre per hour | m/s, km/h |
| Force | newtons, kilonewtons | N, kN |
SI and British Unit Conversions
| Quantity | SI → British | British → SI |
|---|---|---|
| Length | 1 m = 3.281 ft 1 km = 0.621 mile | 1 ft = 0.3048 m 1 mile = 1.609 km |
| Mass | 1 kg = 2.205 lb | 1 lb = 0.454 kg |
| Force | 1 N = 0.225 lbf | 1 lbf = 4.448 N |
| Pressure | 1 N/m² = 0.000145 lbf/in² 1 bar = 14.5038 lbf/in² | 1 lbf/in² = 6.896 kN/m² |
Conversion Problem
A car consumes fuel at the rate of one gallon for a travel of 40 miles. The same car travels a distance of 120 kilometers. What is the consumption of fuel in litres?
Solution:
Step 1: Convert kilometers to miles
1 km = 0.6214 miles → 120 km = 120 × 0.6214 = 74.568 miles
Step 2: Calculate fuel consumption in gallons
Rate: 1 gallon per 40 miles → For 74.568 miles: 74.568/40 = 1.8642 gallons
Step 3: Convert gallons to litres
1 gallon = 4.546 litres → 1.8642 gallons = 1.8642 × 4.546 = 8.475 litres
Final Answer: Fuel consumption = 8.475 litres
4. Units in Measuring Practice with Definitions
| Quantity | Unit | Explanation |
|---|---|---|
| Force (F) | Newton (N) | 1 Newton is equal to the force which imparts an acceleration of 1m/s² to a body of mass 1 kg 1N = 1 kg m/s² |
| Pressure (P) | Pascal (Pa) | 1 Newton per square metre (1 pascal) is equal to the pressure with which the force of 1 N is exercised perpendicular to the area of 1 m² 1Pa = 1 N/m² |
| Work, Energy (W) | Joule (J) | 1 Joule is equal to the work that is done when the point of application of the force of 1 N is shifted by 1 m in the direction of the force 1 J = 1 Nm = 1 Ws = 1 kgm²/s² |
| Power (P) | Watt (W) | 1 Watt is equal to the power with which the energy of 1 J is converted during the time of 1s 1 W = 1 J/s = 1 Nm/s = 1 VA |
Geometrical Quantities
| Symbol | Physical quantity | S.I. Units |
|---|---|---|
| l | Length | Metre (m) |
| A(S) | Area | Square metre (m²) |
| V(v) | Volume | Cubic metre (m³) |
| α,β,γ | Angle | Radian (rad) |
5. Temperature Scales
| Scale | Freezing point | Boiling point |
|---|---|---|
| Centigrade (°C) | 0°C | 100°C |
| Fahrenheit (°F) | 32°F | 212°F |
| Kelvin (K) | 273K | 373K |
Conversion Formulas
°F = (°C × 9/5) + 32
K = °C + 273.15
Temperature Conversion Problem
Convert 120°C to Fahrenheit and Kelvin scales.
Solution:
To Fahrenheit:
°F = (120 × 9/5) + 32 = 216 + 32 = 248°F
To Kelvin:
K = 120 + 273.15 = 393.15K
6. Force in Different Systems
| System | Force Formula | Unit | Definition |
|---|---|---|---|
| C.G.S | Force = Mass × Acceleration | Dyne (dyn) | 1 dyn = 1 gm × 1 cm/sec² |
| F.P.S | Force = Mass × Acceleration | Poundal | 1 poundal = 1 lb × 1 ft/sec² |
| M.K.S | Force = Mass × Acceleration | Newton (N) | 1 N = 1 kg × 1 m/sec² = 10⁵ dynes |
Weight Conversions:
- 1 gm weight = 981 dynes
- 1 lb weight = 32 poundals
- 1 kg weight = 9.81 Newtons
7. Electrical Quantities
| Quantity | Unit | Symbol | Definition |
|---|---|---|---|
| Electric current | Ampere | A | 1 Ampere is the strength of a current which would bring about an electrodynamic force of 0.2×10⁻⁶ N per 1 m length between two parallel conductors placed at a distance of 1 m |
| Electric voltage | Volt | V | 1 Volt is equal to the electric voltage between two points of a metallic conductor in which a power of 1 W is expended for a current of 1 A strength |
| Electric resistance | Ohm | Ω | 1 Ohm is equal to the electric resistance between two points of a metallic conductor in which an electric current of 1 A flows at a voltage of 1 V |
8. Practical Conversion Factors
| Conversion | Factor |
|---|---|
| 1 inch → mm | 1 inch = 25.4 mm |
| 1 mm → inch | 1 mm = 0.03937 inch |
| 1 km → miles | 1 km = 0.621 miles |
| 1 pound → grams | 1 pound = 453.6 g |
| 1 kg → pounds | 1 kg = 2.205 lbs |
| 1 cubic foot → litres | 1 ft³ = 28.317 litres |
9. Units of Physical Quantities
Length Units
| Unit | Conversion |
|---|---|
| Millimetre (mm) | 1 mm = 1000 μm |
| Centimetre (cm) | 1 cm = 10 mm |
| Metre (m) | 1 m = 100 cm |
| Kilometre (km) | 1 km = 1000 m |
| Inch (“) | 1″ = 25.4 mm |
Area Units
| Unit | Conversion |
|---|---|
| Square centimetre (cm²) | 1 cm² = 100 mm² |
| Square metre (m²) | 1 m² = 10,000 cm² |
| Hectare (ha) | 1 ha = 10,000 m² |
| Square inch (sq.in) | 1 sq.in = 6.45 cm² |
10. Angle Measurement Systems
Centesimal System
- 1 Right Angle = 100 grade (100ᵍ)
- 1 grade (1ᵍ) = 100 minutes (100′)
- 1 minute (1′) = 100 seconds (100″)
Sexagesimal System
- 1 Right angle = 90 Degrees (90°)
- 1 Degree (1°) = 60 minutes (60′)
- 1 minute (1′) = 60 seconds (60″)
Circular System (Radians)
1 Radian = 180°/π ≈ 57.3°
1° = π/180 Radians ≈ 0.01745 Radians
Slove problems:
Unit Conversion Problems
Problem 1: Length Conversions
a) 5 yards into metres
1 yard = 0.9144 metres
5 yards × 0.9144 metres/yard = 4.572 metres
b) 15 miles into kilometres
1 mile = 1.60934 kilometres
15 miles × 1.60934 km/mile = 24.1401 kilometres
c) 7 metres into yards
1 metre = 1.09361 yards
7 metres × 1.09361 yards/metre = 7.65527 yards
d) 320 kilometres into miles
1 kilometre = 0.621371 miles
320 km × 0.621371 miles/km = 198.83872 miles
Problem 2: Weight Conversions
a) 5 pounds into kilograms
1 pound = 0.453592 kilograms
5 lbs × 0.453592 kg/lb = 2.26796 kilograms
b) 8.5 kilograms into pounds
1 kilogram = 2.20462 pounds
8.5 kg × 2.20462 lbs/kg = 18.73927 pounds
c) 5 ounces into grams
1 ounce = 28.3495 grams
5 oz × 28.3495 g/oz = 141.7475 grams
d) 16 tons into kilograms
1 ton = 1016.05 kilograms (UK) or 907.185 kg (US)
Assuming US tons: 16 tons × 907.185 kg/ton = 14,514.96 kilograms
Problem 3: Length Conversions (Alternative Units)
a) 40 inches into centimetres
1 inch = 2.54 centimetres
40 inches × 2.54 cm/inch = 101.6 centimetres
b) 12 feet into metres
1 foot = 0.3048 metres
12 feet × 0.3048 m/foot = 3.6576 metres
c) 5 metres into inches
1 metre = 39.3701 inches
5 metres × 39.3701 inches/metre = 196.8505 inches
d) 8 metres into feet
1 metre = 3.28084 feet
8 metres × 3.28084 feet/metre = 26.24672 feet
Problem 4: Volume Conversions
a) 234 cubic metres into gallons
1 cubic metre = 264.172 gallons (US)
234 m³ × 264.172 gallons/m³ = 61,816.248 gallons
b) 2 cubic feet into litres
1 cubic foot = 28.3168 litres
2 ft³ × 28.3168 litres/ft³ = 56.6336 litres
c) 2.5 gallons into litres
1 gallon = 3.78541 litres
2.5 gallons × 3.78541 litres/gallon = 9.463525 litres
d) 5 litres into gallons
1 litre = 0.264172 gallons
5 litres × 0.264172 gallons/litre = 1.32086 gallons
Problem 5: Miscellaneous Conversions
a) 120°C into °F
°F = (°C × 9/5) + 32
(120 × 9/5) + 32 = 216 + 32 = 248°F
b) 8 mm into inches
1 mm = 0.0393701 inches
8 mm × 0.0393701 inches/mm = 0.3149608 inches
c) 12 mm into inches
1 mm = 0.0393701 inches
12 mm × 0.0393701 inches/mm = 0.4724412 inches
Unit Conversion Problems and Solutions
Problem 6: Fuel Consumption Conversion
A car consumes fuel at the rate of one gallon for a travel of 40 miles. The same car travels a distance of 120 kilometer. What is the consumption of fuel in litres?
Solution:
To solve this problem, we need to perform several conversions:
- Convert kilometers to miles to find the equivalent distance in the original units
- Calculate fuel consumption in gallons
- Convert gallons to litres
Step 1: Convert 120 km to miles
1 mile = 1.60934 km
120 km ÷ 1.60934 km/mile ≈ 74.5645 miles
Step 2: Calculate fuel consumption in gallons
The car consumes 1 gallon per 40 miles
Fuel used = 74.5645 miles ÷ 40 miles/gallon ≈ 1.8641 gallons
Step 3: Convert gallons to litres
1 gallon ≈ 3.78541 litres
1.8641 gallons × 3.78541 litres/gallon ≈ 7.0548 litres
Final Answer: The car consumes approximately 7.05 litres of fuel for a 120 km journey.
Problem 7: Equivalent British Units
Write equivalent British units for the given metric units:
Solution:
| Metric Unit | British (Imperial) Unit |
|---|---|
| Seconds, minutes, Hours | Same units (seconds, minutes, hours) – time units are the same in both systems |
| Grams, Kilograms | Ounces (oz), Pounds (lb)
|
| Litres, Cubic meters | Pints, Gallons, Cubic feet
|
| Square centimeter, Square kilometer | Square inch, Square mile
|
Problem 8: Abbreviation Expansion
Expand the abbreviations of the following:
Solution:
| Abbreviation | Expansion | Explanation |
|---|---|---|
| km/l | Kilometers per litre | A measure of fuel efficiency indicating how many kilometers a vehicle can travel using one litre of fuel |
| N/m2 | Newtons per square meter | Unit of pressure, equivalent to Pascal (Pa) |
| KW | Kilowatt | Unit of power equal to 1000 watts |
| m/s2 | Meters per second squared | Unit of acceleration – change in velocity (m/s) per second |
| RPM | Revolutions per minute | A measure of rotational speed – number of complete turns in one minute |
Workshop Calculation & Science (NSQF) – Units
Assignment – Conversions of length, mass, force, work, power and energy
a. Length Conversions
i) 3.4 m = ______ mm
Solution:
1 meter = 1000 millimeters (mm)
3.4 m = 3.4 × 1000 mm
3.4 m = 3400 mm
ii) 1.2 m = ______ cm
Solution:
1 meter = 100 centimeters (cm)
1.2 m = 1.2 × 100 cm
1.2 m = 120 cm
iii) 0.8 m = ______ mm
Solution:
1 meter = 1000 millimeters (mm)
0.8 m = 0.8 × 1000 mm
0.8 m = 800 mm
iv) 0.02 km = ______ cm
Solution:
1 kilometer = 1000 meters
1 meter = 100 centimeters
So, 1 km = 1000 × 100 = 100,000 cm
0.02 km = 0.02 × 100,000 cm
0.02 km = 2000 cm
v) 10.2 km = ______ mile
Solution:
1 kilometer ≈ 0.621371 miles
10.2 km = 10.2 × 0.621371 miles
10.2 km ≈ 6.338 miles (rounded to 3 decimal places)
vi) 6 m = ______ km
Solution:
1 kilometer = 1000 meters
6 m = 6 ÷ 1000 km
6 m = 0.006 km
vii) 18 m = ______ mm
Solution:
1 meter = 1000 millimeters
18 m = 18 × 1000 mm
18 m = 18000 mm
viii) 450 m = ______ km
Solution:
1 kilometer = 1000 meters
450 m = 450 ÷ 1000 km
450 m = 0.45 km
ix) 85 cm = ______ km
Solution:
1 kilometer = 1000 meters
1 meter = 100 centimeters
So, 1 km = 1000 × 100 = 100,000 cm
85 cm = 85 ÷ 100,000 km
85 cm = 0.00085 km or 8.5 × 10⁻⁴ km
x) 0.06 km = ______ mm
Solution:
1 kilometer = 1000 meters
1 meter = 1000 millimeters
So, 1 km = 1000 × 1000 = 1,000,000 mm
0.06 km = 0.06 × 1,000,000 mm
0.06 km = 60000 mm
b. Mass Conversions
i) 650 g = ______ kg
Solution:
1 kilogram = 1000 grams
650 g = 650 ÷ 1000 kg
650 g = 0.65 kg
ii) 300 cg = ______ g
Solution:
1 centigram = 0.01 grams
300 cg = 300 × 0.01 g
300 cg = 3 g
iii) 8 g = ______ dg
Solution:
1 decigram = 0.1 grams
Therefore, 1 g = 10 dg
8 g = 8 × 10 dg
8 g = 80 dg
iv) 120 mg = ______ g
Solution:
1 milligram = 0.001 grams
120 mg = 120 × 0.001 g
120 mg = 0.12 g
v) 8 dag = ______ mg
Solution:
1 decagram = 10 grams
1 gram = 1000 milligrams
So, 1 dag = 10 × 1000 = 10,000 mg
8 dag = 8 × 10,000 mg
8 dag = 80,000 mg
vi) 2.5 g = ______ mg
Solution:
1 gram = 1000 milligrams
2.5 g = 2.5 × 1000 mg
2.5 g = 2500 mg
vii) 2.5 g = ______ kg
Solution:
1 kilogram = 1000 grams
2.5 g = 2.5 ÷ 1000 kg
2.5 g = 0.0025 kg
viii) 20 cg = ______ mg
Solution:
1 centigram = 10 milligrams
20 cg = 20 × 10 mg
20 cg = 200 mg
ix) 0.05 Mt = ______ kg
Solution:
Assuming Mt stands for metric tons (tonnes)
1 metric ton = 1000 kilograms
0.05 Mt = 0.05 × 1000 kg
0.05 Mt = 50 kg
c. Force Conversions
i) 1.2 N = ______ kg
Solution:
This conversion assumes standard gravity (9.80665 m/s²)
F = m × a ⇒ m = F/a
1 N = 1 kg·m/s²
1.2 N = 1.2 ÷ 9.80665 kg (force)
1.2 N ≈ 0.122 kg (rounded to 3 decimal places)
ii) 2.6 N = ______ kg
Solution:
Using standard gravity (9.80665 m/s²)
2.6 N = 2.6 ÷ 9.80665 kg (force)
2.6 N ≈ 0.265 kg (rounded to 3 decimal places)
iii) 800 N = ______ KN
Solution:
1 kilonewton = 1000 newtons
800 N = 800 ÷ 1000 kN
800 N = 0.8 kN
iv) 14.5 kg = ______ N
Solution:
Using standard gravity (9.80665 m/s²)
F = m × a
14.5 kg × 9.80665 m/s² ≈ 142.196 N (rounded to 3 decimal places)
v) 25 kg = ______ N
Solution:
Using standard gravity (9.80665 m/s²)
25 kg × 9.80665 m/s² ≈ 245.166 N (rounded to 3 decimal places)
d. Work, Energy, Amount of Heat Conversions
i) 2 Nm = ______ Ncm
Solution:
1 meter = 100 centimeters
2 Nm = 2 × 100 Ncm
2 Nm = 200 Ncm
ii) 50 Ncm = ______ Nm
Solution:
1 meter = 100 centimeters
50 Ncm = 50 ÷ 100 Nm
50 Ncm = 0.5 Nm
iii) 120 KJ = ______ J
Solution:
1 kilojoule = 1000 joules
120 kJ = 120 × 1000 J
120 kJ = 120,000 J
iv) 40 J = ______ KJ
Solution:
1 kilojoule = 1000 joules
40 J = 40 ÷ 1000 kJ
40 J = 0.04 kJ
v) 300 wh = ______ kwh
Solution:
1 kilowatt-hour = 1000 watt-hours
300 wh = 300 ÷ 1000 kWh
300 wh = 0.3 kWh
e. Power Conversions
i) 200 mW = ______ W
Solution:
1 milliwatt = 0.001 watts
200 mW = 200 × 0.001 W
200 mW = 0.2 W
ii) 0.2 kW = ______ W
Solution:
1 kilowatt = 1000 watts
0.2 kW = 0.2 × 1000 W
0.2 kW = 200 W
iii) 300 kW = ______ mW
Solution:
1 kilowatt = 1000 watts
1 watt = 1000 milliwatts
So, 1 kW = 1000 × 1000 = 1,000,000 mW
300 kW = 300 × 1,000,000 mW
300 kW = 300,000,000 mW or 3 × 10⁸ mW
iv) 2.106 W = ______ mW
Solution:
1 watt = 1000 milliwatts
2.106 W = 2.106 × 1000 mW
2.106 W = 2106 mW
v) 6.10⁻⁴ kW = ______ W
Solution:
1 kilowatt = 1000 watts
6.10 × 10⁻⁴ kW = 6.10 × 10⁻⁴ × 1000 W
6.10 × 10⁻⁴ kW = 0.61 W
vi) 2 W = ______ KW
Solution:
1 kilowatt = 1000 watts
2 W = 2 ÷ 1000 kW
2 W = 0.002 kW
vii) 350 W = ______ kW
Solution:
1 kilowatt = 1000 watts
350 W = 350 ÷ 1000 kW
350 W = 0.35 kW
viii) 0.08 W = ______ kW
Solution:
1 kilowatt = 1000 watts
0.08 W = 0.08 ÷ 1000 kW
0.08 W = 0.00008 kW or 8 × 10⁻⁵ kW
ix) 2 × 10⁻³ kW = ______ W
Solution:
1 kilowatt = 1000 watts
2 × 10⁻³ kW = 2 × 10⁻³ × 1000 W
2 × 10⁻³ kW = 2 W
x) 0.04 W = ______ mW
Solution:
1 watt = 1000 milliwatts
0.04 W = 0.04 × 1000 mW
0.04 W = 40 mW
f. Miscellaneous Conversions
i) 3 Nm = ______ J
Solution:
1 newton-meter = 1 joule
3 Nm = 3 J
ii) 2 J = ______ Ws
Solution:
1 joule = 1 watt-second
2 J = 2 Ws
iii) 12 J = ______ KJ
Solution:
1 kilojoule = 1000 joules
12 J = 12 ÷ 1000 kJ
12 J = 0.012 kJ
iv) 3 Nm/s = ______ J/s
Solution:
1 Nm = 1 J, so 1 Nm/s = 1 J/s
3 Nm/s = 3 J/s
v) 5 N = ______ KN
Solution:
1 kilonewton = 1000 newtons
5 N = 5 ÷ 1000 kN
5 N = 0.005 kN
vi) 3 KJ = ______ Nm
Solution:
1 kilojoule = 1000 joules
1 joule = 1 newton-meter
3 kJ = 3 × 1000 Nm
3 kJ = 3000 Nm
vii) 18 J/s = ______ W
Solution:
1 watt = 1 joule/second
18 J/s = 18 W
viii) 12 W = ______ J/s
Solution:
1 watt = 1 joule/second
12 W = 12 J/s
ix) kJ/s = ______ Nm/s
Solution:
1 kilojoule = 1000 joules
1 joule = 1 newton-meter
Therefore, 1 kJ/s = 1000 Nm/s
(This is a general conversion, not a specific value)
Unit Conversion
